Question: Let $y=\dfrac{\sqrt{x}}{e^x}$. $\dfrac{dy}{dx}=$
$\dfrac{\sqrt{x}}{e^x}$ is the quotient of two, more basic, expressions: $\sqrt{x}$ and $e^x$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( x √ e x ) = d d x ( x √ ) e x − x √ d d x ( e x ) ( e x ) 2 = 1 2 x √ ⋅ e x − x √ ⋅ e x ( e x ) 2 = e x ( 1 2 x √ − x √ ) ( e x ) 2 = 1 − 2 x 2 x √ e x = 1 − 2 x 2 e x x √ The quotient rule Differentiate x √ and e x Simplify Combine numerator Simplify \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{\sqrt{x}}{e^x}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sqrt{x})e^x-\sqrt{x}\dfrac{d}{dx}(e^x)}{(e^x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac1{2\sqrt{x}}\cdot e^x-\sqrt{x}\cdot e^x}{(e^x)^2}&&\gray{\text{Differentiate }\sqrt{x}\text{ and }e^x} \\\\ &=\dfrac{\cancel{e^x}\left(\dfrac{1}{2\sqrt{x}}-\sqrt{x}\right)}{(e^x)^\cancel{2}}&&\gray{\text{Simplify}} \\\\ &=\dfrac{\dfrac{1-2x}{2\sqrt{x}}}{e^x}&&\gray{\text{Combine numerator}} \\\\ &={\dfrac{1-2x}{2e^x\sqrt{x}}}&&\gray{\text{Simplify}} \end{aligned} In conclusion, $\dfrac{dy}{dx}={\dfrac{1-2x}{2e^x\sqrt{x}}}$ or any other equivalent form.